Dirichlet problem for manifold, how to prove $W^{1,2}_0(\Omega)$ solution is $C^{2,\alpha}(\bar{\Omega})$? Let $M$ be an n dimensional Riemannian manifold without boundary. Let $\Omega \subset M$ be a bounded domain with smooth boundary. Let $f \in C^{\alpha}(\bar{\Omega})$, Consider the Dirichlet problem.
$$
\Delta u=f \ \text{ in }\Omega,
u|_{\partial \Omega}=0.
$$
Do we have a solution $u\in C^{2,\alpha}(\bar{\Omega})$? 
The following argument is inspired by How to solve the $C^\alpha$ Poisson equation on closed Riemannian manifolds?. 
Since $f$ is bounded, $f\in L^2(\Omega)$, by the existence of minizer of the energy functional (just like the Euclidean case), there exists a weak solution $u\in W_0^{1,2}(U)$. For $x\in \Omega$, let $U$ be a small neighborhood contained in a coordinate chart, consider 
$$
\Delta v=f \text{ in }U,
v|_{\partial U}=0.
$$
By the theory of Dirichlet problem for Euclidean space,  we have a solution $v\in C^{2,\alpha}(\bar{U})$. Since 
$$
\Delta (u-v)=0 \text{ in } U,
$$
then $u-v\in C^{\infty}(U)$, so $u\in C^{2,\alpha}(U)$. By the arbitrariness of x, we know $u\in C^{2,\alpha}(\Omega)$. 
But how to prove that $u\in C^{2,\alpha}(\bar{\Omega})$?
 A: I will essentially explain the comment under Theorem 8.14 in Gilbarg-Trudinger. 
I will assume the result stated there: given smooth boundary data and RHS, the Poisson equation has a unique smooth solution. You are already considering the Dirichlet problem, so we don't have to worry about boundary data. 
Your link contains a comment that $C^\infty(\overline\Omega)$ is not dense in $C^{2,\alpha}(\overline\Omega)$, which is true. But it is true that given $f\in C^{2,\alpha}(\overline\Omega)$ and $\beta\in (0,\alpha)$, there is a sequence $f_i\in C^{\infty}(\overline\Omega)$ converging to $f$ in the $C^{2,\beta}(\Omega)$ norm. 
So let $u_i\in C^\infty(\overline\Omega)$ be the Dirichlet solution of $\Delta u_i=f_i$. We can apply Schauder estimates 
$$\|u_i\|_{2,\alpha;\Omega}\le C(\|u_i\|_{0;\Omega}+\|f_i\|_{0,\alpha;\Omega}),\tag{$*$}$$
where $C$ is a constant independent of $i$. By the maximum principle, it is possible to bound
$$\|u_i\|_{0;\Omega}\le C\|f_i\|_{0;\Omega}\le C\|f_i\|_{0,\alpha;\Omega},$$
see Theorem 3.7 in Gilbarg-Trudinger. So $(*)$ becomes
$$\|u_i\|_{2,\alpha;\Omega}\le C\|f_i\|_{0,\alpha;\Omega}.\tag{$**$}$$
Now we can actually choose the $f_i$'s so that $\|f_i\|_{0,\alpha;\Omega}\le 2\|f\|_{0,\alpha;\Omega}$ for any $i$, so $(**)$ just says that $(u_i)$ is bounded in $C^{2,\alpha}(\overline\Omega)$. By the Arzela-Ascoli theorem, we obtain a convergent subsequence to a function $u$ in $C^{2,\beta}(\overline\Omega)$ and since $f_i\to f$ in $C^{0,\beta}(\overline\Omega)$, $\Delta u=f$. 
The key observation now is that even though the convergence of the $u_i$'s is in $C^{2,\beta}(\overline\Omega),$ the function $u$ itself is in $C^{2,\alpha}(\overline\Omega)$. This follows just from the uniform convergence of $u_i$, $\nabla u_i$, and $\nabla^2u_i$, and the fact that these all satisfy the $\alpha$-Hölder condition with the same constants.  
